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STRUCTURE OF Ce-136, Ce-138, Ce-140, Ce-142
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Nuclear structure of Cerium with 28 blank positions Naturally occurring cerium (Ce) is composed of 4 stable isotopes: Ce-136, Ce-138, Ce-140, and Ce-142 with Ce-140 being the most abundant (88.48% natural abundance) and the only one theoretically stable; Ce-136, Ce-138, and Ce-142 are predicted to undergo double beta decay but this process has never been observed. Comparing cerium of 58 protons (even number ) with barium of 56 protons (even number) one sees that the structure of cerium is also of high symmetry. ( See my STRUURE OF Ba-132...Ba-138 ). In the following diagram of cerium the additional p57n57 and p58n58 are not shown, but you can see the symmetrical positions of n57 and p58 by using the top view of the third horizontal plane. Note that p53, n54, n55, p56, n57 and p58 fill the blank positions of (p) and (n). Thus, the third and the fourth plane have 2(n) +2(n) blank positions for receiving 4 extra neutrons. As in the case of Ba here the other horizontal planes along the two squares have the same number of blank positions given by The two squares give 8n The first and sixth plane give 4(n) The second and fifth plane give 4{n} +8n The third and the fourth plane give 4(n) That is N = 8n + 4(n) + 4{n} + 8n + 4(n) = 28 blank positions able to receive extra neutrons of opposite spins STRUCTURE OF Ce-136, Ce-138, Ce,140 AND Ce-142 WITH S =0 Since the 58 protons and 58 neutrons of Cerium give S=0 we conclude that the total spin S=0 of the above nuclides is due to the extra neutrons of opposite spins. For example the Ce-136 of 20 extra neutrons has 4{n} + 16n of opposite spins, while the Ce-142 of 26 extra neutrons has 4{n} + 16n + 6(n) of opposite spins. DIAGRAM OF CERIUM FORMING 28 BLANK POSITIONS Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 and p50n50 which makes the two symmetrical vertical rectangles with the n15p17 and the p16n18 respectively. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the 4 extra neutrons (n) of the first and the sixth plane are not shown, while the extra neutrons 4n existing over the p31 and p32, and under the p21 and p22 along with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.......p40.......n' ' n........p38..........n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n.......p37......n37 ' ' n39.....p39..........n ' H. Square with n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS Here the p53, n54, n55, p56, n57, and p58 fill the blank positions of (n) and (p). Thus the plane gives also 4 blank positions to receive 2(n) and 2(p) (n)..........p58........n50.......p51.......(n ) p53....... n42........p16......n16......p44........n54 n47........p25........n6........p6........n26.........p48 p45........n25........p5........n5........p26........ n46 n55........p41.......n15.......p15.......n43.......p56 ' n57........p49.......n52' Category:Fundamental physics concepts